# The value of $\sin {20^0}\sin {40^0}\sin {60^0}\sin {80^0}$ is

$\left( A \right)$. $\dfrac{{ - 3}}{{\sqrt {16} }}$

\[\left( B \right)\]. $\dfrac{5}{{\sqrt {16} }}$

$\left( C \right)$. $\dfrac{3}{{\sqrt {16} }}$

$\left( D \right)$. $\dfrac{{ - 5}}{{\sqrt {16} }}$

Answer

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Hint: In the above problem the trigonometric product to sum identities should be used. The identity to be used is to be determined by inspection so that the result gives angle with general known trigonometric ratio.

Given in the problem, we need to find value of the expression

$\sin {20^0}\sin {40^0}\sin {60^0}\sin {80^0}$ …………………………….. (1)

We need to group terms so that using a product to sum trigonometric formula gives angles whose trigonometric value is known.

We know that value of

$\cos (x - y) - \cos (x + y) = 2\sin x\sin y$ …………………………………...(2)

Put $x = 80$and $y = 40$in the equation (2) ,we get

\[

2\sin {80^0}\sin {40^0} = \cos ({80^0} - {40^0}) - \cos ({80^0} + {40^0}) \\

\Rightarrow 2\sin {80^0}\sin {40^0} = \cos ({40^0}) - \cos ({120^0}) \\

\]

We know that $\cos \left( {{{180}^0} - \theta } \right) = - \cos \theta $

Put $\theta = {60^0}$ in above $ \Rightarrow \cos {120^0} = \cos \left( {{{180}^0} - {{60}^0}} \right) = - \cos {60^0} = - \dfrac{1}{2}$

\[ \Rightarrow 2\sin {80^0}\sin {40^0} = \cos ({40^0}) - \cos ({120^0}) = \cos ({40^0}) + \dfrac{1}{2}\] …………………(3)

Multiplying expression (1) with $\dfrac{2}{2}$ and rearranging, we get

$\dfrac{{2\sin {{80}^0}\sin {{40}^0}\sin {{60}^0}\sin {{20}^0}}}{2}$

Using equation (3) in above, we get

$\left( {\cos {{40}^0} + \dfrac{1}{2}} \right)\dfrac{{\sin {{60}^0}\sin {{20}^0}}}{2}$

Using $\sin {60^0} = \dfrac{{\sqrt 3 }}{2}$ in above,

$ \Rightarrow \left( {\cos {{40}^0} + \dfrac{1}{2}} \right)\dfrac{{\sqrt 3 }}{4}\sin {20^0}$

$ \Rightarrow \dfrac{{\sqrt 3 }}{8}\left( {2\sin {{20}^0}\cos {{40}^0} + \sin {{20}^0}} \right)$ …………………………………………………...(4)

We know that value of

$\sin (x + y) + \sin (x - y) = 2\sin x\cos y$ ……………………………………………...(5)

Put $x = 20$and $y = 40$in the equation (5), we get

\[

2\sin {20^0}\cos {40^0} = \sin ({20^0} + {40^0}) + \sin ({20^0} - {40^0}) \\

\Rightarrow 2\sin {20^0}\sin {40^0} = \sin ({60^0}) + \sin ( - {20^0}) \\

\]

We know that $\sin \left( { - \theta } \right) = - \sin \theta $

Put $\theta = {20^0}$ in above gives $\sin \left( { - {{20}^0}} \right) = - \sin {20^0}$

\[ \Rightarrow 2\sin {20^0}\sin {40^0} = \sin ({60^0}) - \sin ({20^0}) = \dfrac{{\sqrt 3 }}{2} - \sin ({20^0})\] ……………………………….(6)

Using equation (6) in (4), we get

\[

\dfrac{{\sqrt 3 }}{8}\left( {2\sin {{20}^0}\cos {{40}^0} + \sin {{20}^0}} \right) = \dfrac{{\sqrt 3 }}{8}\left( {\dfrac{{\sqrt 3 }}{2} - \sin {{20}^0} + \sin {{20}^0}} \right) \\

\Rightarrow \dfrac{{\sqrt 3 }}{8}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{3}{{16}} \\

\]

Therefore, the value of expression (1) is $\dfrac{3}{{16}}$.

Hence option $(C)$ $\dfrac{3}{{16}}$ is the correct answer.

Note: Always remember trigonometric sum to product and product to sum formula. Modifications may need to be performed in the expressions like above in order to use these identities. These modifications should never alter the value of the original expression. Try to convert the expression in problems of above type into known trigonometric ratio values.

Given in the problem, we need to find value of the expression

$\sin {20^0}\sin {40^0}\sin {60^0}\sin {80^0}$ …………………………….. (1)

We need to group terms so that using a product to sum trigonometric formula gives angles whose trigonometric value is known.

We know that value of

$\cos (x - y) - \cos (x + y) = 2\sin x\sin y$ …………………………………...(2)

Put $x = 80$and $y = 40$in the equation (2) ,we get

\[

2\sin {80^0}\sin {40^0} = \cos ({80^0} - {40^0}) - \cos ({80^0} + {40^0}) \\

\Rightarrow 2\sin {80^0}\sin {40^0} = \cos ({40^0}) - \cos ({120^0}) \\

\]

We know that $\cos \left( {{{180}^0} - \theta } \right) = - \cos \theta $

Put $\theta = {60^0}$ in above $ \Rightarrow \cos {120^0} = \cos \left( {{{180}^0} - {{60}^0}} \right) = - \cos {60^0} = - \dfrac{1}{2}$

\[ \Rightarrow 2\sin {80^0}\sin {40^0} = \cos ({40^0}) - \cos ({120^0}) = \cos ({40^0}) + \dfrac{1}{2}\] …………………(3)

Multiplying expression (1) with $\dfrac{2}{2}$ and rearranging, we get

$\dfrac{{2\sin {{80}^0}\sin {{40}^0}\sin {{60}^0}\sin {{20}^0}}}{2}$

Using equation (3) in above, we get

$\left( {\cos {{40}^0} + \dfrac{1}{2}} \right)\dfrac{{\sin {{60}^0}\sin {{20}^0}}}{2}$

Using $\sin {60^0} = \dfrac{{\sqrt 3 }}{2}$ in above,

$ \Rightarrow \left( {\cos {{40}^0} + \dfrac{1}{2}} \right)\dfrac{{\sqrt 3 }}{4}\sin {20^0}$

$ \Rightarrow \dfrac{{\sqrt 3 }}{8}\left( {2\sin {{20}^0}\cos {{40}^0} + \sin {{20}^0}} \right)$ …………………………………………………...(4)

We know that value of

$\sin (x + y) + \sin (x - y) = 2\sin x\cos y$ ……………………………………………...(5)

Put $x = 20$and $y = 40$in the equation (5), we get

\[

2\sin {20^0}\cos {40^0} = \sin ({20^0} + {40^0}) + \sin ({20^0} - {40^0}) \\

\Rightarrow 2\sin {20^0}\sin {40^0} = \sin ({60^0}) + \sin ( - {20^0}) \\

\]

We know that $\sin \left( { - \theta } \right) = - \sin \theta $

Put $\theta = {20^0}$ in above gives $\sin \left( { - {{20}^0}} \right) = - \sin {20^0}$

\[ \Rightarrow 2\sin {20^0}\sin {40^0} = \sin ({60^0}) - \sin ({20^0}) = \dfrac{{\sqrt 3 }}{2} - \sin ({20^0})\] ……………………………….(6)

Using equation (6) in (4), we get

\[

\dfrac{{\sqrt 3 }}{8}\left( {2\sin {{20}^0}\cos {{40}^0} + \sin {{20}^0}} \right) = \dfrac{{\sqrt 3 }}{8}\left( {\dfrac{{\sqrt 3 }}{2} - \sin {{20}^0} + \sin {{20}^0}} \right) \\

\Rightarrow \dfrac{{\sqrt 3 }}{8}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{3}{{16}} \\

\]

Therefore, the value of expression (1) is $\dfrac{3}{{16}}$.

Hence option $(C)$ $\dfrac{3}{{16}}$ is the correct answer.

Note: Always remember trigonometric sum to product and product to sum formula. Modifications may need to be performed in the expressions like above in order to use these identities. These modifications should never alter the value of the original expression. Try to convert the expression in problems of above type into known trigonometric ratio values.

Last updated date: 24th Sep 2023

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